Mathematics - Differential Equation Question with Solution | TestHub

Put $\mathrm{y}=\mathrm{vx}\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$ D.E. $\begin{array}{rl}& {\mathrm{x}}^2\left(v+\mathrm{x}\frac{\mathrm{d}v}{\mathrm{dx}}\right)+{\mathrm{x}}^{2}v={\mathrm{x}}^2\left(1+v^2\right)\\ & \Rightarrow v+x\frac{dv}{dx}+v=1+v^2\\ & \Rightarrow x\frac{dv}{dx}=1+v^2-2v\\ & \Rightarrow \int \frac{dv}{(v-1{)}^2}=\int \frac{dx}{x}\\ & \Rightarrow -\frac{1}{v-1}=\ln |x|+\mathrm{C}\end{array}$ $\Rightarrow \frac{\mathrm{x}}{\mathrm{x}-\mathrm{y}}=\ln |x|+\mathrm{C}=\ln \mathrm{x}+\mathrm{C}($ Since $\mathrm{x}>\frac{1}{\mathrm{e}})$ Given $y(1)=0$ $\Rightarrow C=1$ So $\frac{x}{x-y}=\ln (ex)$ Now $y(e)=\frac{e}{2}$ and $y\left(e^2\right)=\frac{2e^2}{3}$ $\therefore \frac{2(y(e){)}^2}{y\left(e^2\right)}=\frac{2\cdot \frac{e^2}{4}}{\frac{2e^2}{3}}=\frac{3}{4}=0.75$